For what values of x is #f(x)= x^4-3x^3-4x-7 # concave or convex?

Question

For what values of x is #f(x)= x^4-3x^3-4x-7  # concave or convex?

Isaiah Allen · Accepted Answer

the function is convex in #]-oo;0[uu]2/3;+oo[# and concave in #]0;2/3[#

You would analyze the second derivative; the first one is:

#f'(x)=4x^3-9x^2-4#

then the second one is:

#f''(x)=12x^2-18x#

Let's solve the inequality #f''(x)>0#:

#12x^2-8x>0#

that's

#x<0 or x>2/3#

Then the given function is convex in

#]-oo;0[uu]2/3;+oo[#

and concave in

#]0;2/3[#

graph{x^4-3x^3-4x-7 [-5, 5, -27, 10]}

Charles Anctil · Answer

undefined To determine where the function $ f(x) = x^4 - 3x^3 - 4x - 7 $ is concave or convex, you need to find the second derivative of the function and then analyze its sign.

First, find the second derivative:
$$ f''(x) = 12x^2 - 18x $$

Next, set $ f''(x) $ equal to zero and solve for $ x $:
$$ 12x^2 - 18x = 0 $$
$$ 6x(2x - 3) = 0 $$
$$ x = 0 \quad \text{or} \quad x = \frac{3}{2} $$

Now, you can test intervals created by these critical points with the second derivative test:
1. Choose a value of $ x $ less than $ 0 $ (e.g., $ x = -1 $) and plug it into $ f''(x) $ to determine its sign.
2. Choose a value of $ x $ between $ 0 $ and $ \frac{3}{2} $ (e.g., $ x = 1 $) and plug it into $ f''(x) $.
3. Choose a value of $ x $ greater than $ \frac{3}{2} $ (e.g., $ x = 2 $) and plug it into $ f''(x) $.

Analyzing the signs of $ f''(x) $ in these intervals will determine where the function is concave up (convex) or concave down.

Charles Arocha · Answer

undefined The function f(x) = x^4 - 3x^3 - 4x - 7 is concave upward for x < 0 and concave downward for x > 0.